Geometric Problems

Sequences have wide applications in geometric problems, from simple figure dissection to complex fractal geometry.

Figure Dissection Problems

Example 1: Triangle Dissection

Connect the midpoints of the three sides of a triangle to obtain 4 small triangles. Repeat this process. How many triangles are there after the nnth dissection?

Solution:

  • 1st time: 4
  • 2nd time: 4×4=164 \times 4 = 16
  • 3rd time: 4×4×4=644 \times 4 \times 4 = 64
  • nnth time: 4n4^n

This is a geometric sequence with common ratio 4.

Example 2: Square Dissection

Divide a square into four equal parts, remove the upper-right part, and repeat this process on the remaining three squares. What is the remaining area after the nnth step?

Solution:

Let the area of the original square be 1. Each time 34\frac{3}{4} remains:

Sn=(34)nS_n = \left(\frac{3}{4}\right)^n

This is a geometric sequence, and limnSn=0\lim_{n \to \infty} S_n = 0

Fractal Geometry

The Sierpinski Triangle

Start with an equilateral triangle and remove the small middle triangle each time, repeating infinitely.

After the nnth step:

  • Number of triangles: 3n3^n
  • Total area: (34)nS0\left(\frac{3}{4}\right)^n S_0 (S0S_0 is the original area)

As nn \to \infty:

  • The number of triangles \to \infty
  • The total area 0\to 0

This is the wonder of fractals!

Infinite Series and Area

Example: An Infinitely Divided Square

A square with side length 1. Take away 12,14,18,\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \ldots of its area in turn. What is the remaining area?

Solution:

The sum of the removed areas:

S=12+14+18+=12112=1S = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = 1

Remaining area: 11=01 - 1 = 0

Solid Geometry

Example: Stacking Balls

Layer 1 has 1 ball, layer 2 has 4, layer 3 has 9, …, layer nn has n2n^2 balls. How many balls are there in the first nn layers in total?

Solution:

Sn=12+22+32++n2=n(n+1)(2n+1)6S_n = 1^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}

Practice Problems

Exercise 1

Divide a square with side length 1 into 4 small squares, then divide each small square into 4 even smaller squares, repeating this nn times. How many small squares are there after the nnth step? What is the side length of each small square?

Reference Answer(2 个标签)
sequence applicationgeometric problem

Solution:

Number of small squares: 4n4^n

Side length of each small square: (12)n=12n\left(\frac{1}{2}\right)^n = \frac{1}{2^n}

Answer: 4n4^n squares, with side length 12n\frac{1}{2^n}

Exercise 2

An equilateral triangle with side length 1. Connect the midpoints of the sides to obtain 4 small triangles and remove the middle one. Repeat this operation on the remaining 3 triangles. Find the perimeter of the remaining figure after the nnth operation.

Reference Answer(2 个标签)
sequence applicationgeometric problem

Solution:

Initial perimeter: 3

1st time: after removing the middle triangle, the perimeter becomes 3×32=4.53 \times \frac{3}{2} = 4.5

2nd time: each small triangle’s perimeter becomes 32\frac{3}{2} times the original

Perimeter after the nnth time: 3×(32)n3 \times \left(\frac{3}{2}\right)^n

Answer: 3×(32)n3 \times \left(\frac{3}{2}\right)^n

Note: the perimeter tends to infinity, but the area tends to 0!


Summary

Symbols Used in This Article

符号类型读音/说明在本文中的含义
nn数学符号nNumber of dissections
SnS_n数学符号S-sub-nThe area after the nnth step
S0S_0数学符号S-sub-zeroThe initial area
34\frac{3}{4}数学符号three quartersThe proportion of side remaining each time
\infty数学符号infinityInfinity
limn\lim_{n \to \infty}数学符号limit as n approaches infinityThe limit as nn tends to infinity
SS数学符号SThe sum of an infinite series

中英对照

中文术语英文术语音标说明
分形fractal/ˈfræktəl/A self-similar geometric figure
谢尔宾斯基三角形Sierpinski triangle/sɪərˈpɪnski ˈtraɪæŋɡəl/A classic fractal figure
无穷级数infinite series/ˈɪnfɪnət ˈsɪəriːz/The sum of infinitely many terms
等比数列geometric sequence/ˌdʒiːəˈmetrɪk ˈsiːkwəns/A sequence with a fixed common ratio
周长perimeter/pəˈrɪmɪtə/The length of a figure’s boundary